Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the probability density function (PDF) of $f(x)=1/\sin(x)$ when $x$ is uniformly distributed in $(0,90)$?

$f(x)=\sin(x)$ has a known PDF, which has the form $2(\pi\sqrt{1-\sin(x)^2})^{-1}$, but I cannot find the PDF for $1/\sin(x)$. The latter would have interesting applications in astronomy, specially for so-called "luminosity functions".

Thank you very much


share|cite|improve this question

1 Answer 1

up vote 4 down vote accepted

Let $z=1/\sin(x)$, then $x=\sin^{-1}(1/z)$ on $(0,\frac{\pi}{2})$ interval. The pdf you seek, is the differential of uniform c.d.f., which is $\frac{2}{\pi} dx$

$$ \frac{2}{\pi} d( \sin^{-1}( z^{-1} ) ) = \frac{2}{\pi z} \frac{1}{\sqrt{z^2-1}},$$ where variable $z>1$.

By the way, you could have used Mathematica to directly find this:

In[100]:= FullSimplify[PDF[TransformedDistribution[1/Sin[x], 
       x \[Distributed] UniformDistribution[{0, Pi/2}]], z]]

Out[100]= Piecewise[{{2/(Pi*z*Sqrt[-1 + z^2]), z > 1}}, 0]
share|cite|improve this answer
Super. Thank you very much, Sasha. – Sebastian Aug 4 '11 at 16:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.